a good quesn

before collison:
v_a=√2ga
v_b=0

after collisoin
v_a=(m-2m/3)√2ga/(m+2m)

now calculate height

See

F = kx for both

Acceleration for 1 will be kx/m₁
Acceleration for 2 will be kx/m₂

S₁ = kx*t²/2m₁
S₂ = kx*t²/2m₂
W₁/W₂= F * S₁ / F * S₂
W₁/W₂ = kx*t²2m₂/2m₁kx*t².
Therefore

W₁/W₂ = m₂/m₁

Is my answer rite?

Nishant Bhaiyya pls check my solution

give d solution i need for dees question im confused

but how is W1/W2=m2/m1 plz xplain & even how is VA-VB=ERteta it isnt mentioned dat teta is small xplain

W₁/W₂ = m₂/m₁

Conceptual question

Good one

ans for post #5 part 2
mgh = mv²/2
v is the velocity when it hit the ball
v=√2g*a
e = 1/3
e = v/v₂-v₁
v₁ velocity of 2m v₂ is velocity of m
mv = 2mv₁ +mv₂
v₂ = v - 2v₁
1/3 = v - v/3v₁
v - 3v₁= 3v
3v₁ = 4v
v₁= 2√2ga/3
2mgh = 1/2mv²
2g*h = 4*2*g*a/9*2
h =2a/9

that one was for post#5 second part

a straight uniform meal rod of length 3l is bent through right angle as shown.The bent rod is den placed on a rough horizontal table.a light string is attached to d vertex of d right angle. d string is den pulled horizontally so dat d rod translates at constant velocity den d angle alpha which d side 2l makes with string is:

ans for post#7
V_a-V_b=W
=>V-V_b=E(Rθ)
=>V_b=V-ERθ

nish byah has already given the hint ..............

plz reply fast i need a complete solution if possible or else give at least hint which is useful

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